Why Put Random Effects on Both Parameters?
In a population model, two subjects exposed to the same concentration need not eliminate drug at the same rate. With Michaelis–Menten elimination, that rate depends on both \(V_{\max}\), the maximum elimination rate, and \(K_m\), the concentration at which the rate is half of \(V_{\max}\). Interindividual differences in either parameter can therefore change a subject’s concentration-time profile, and differences in the two parameters can reinforce or offset one another.
The earlier material on parameterizing \(V_{\max}\) and \(K_m\) in population models establishes the role of typical parameter values. Here, random effects specify how an individual subject’s parameters vary around those values. The focus is on the consequences of that specification, not on fitting algorithms or on correlations between the two effects.
The exponential transformation makes both individual parameters positive whenever the typical values are positive. The random effects themselves are dimensionless: they act on the log scale, while \(V_{\max,i}\) and \(TVV_{\max}\) share units (for example, mg/h), as do \(K_{m,i}\) and \(TVK_m\) (for example, mg/L). A positive \(\eta_{V,i}\) raises subject \(i\)’s \(V_{\max}\) above the typical value; a negative one lowers it. The same direction applies to \(\eta_{K,i}\) and \(K_m\).
The effects describe between-subject parameter differences in this model. They are not residual measurement errors, and they do not by themselves imply that a subject’s parameters change over time. If a model includes covariates or time-varying parameters, those are additional parts of the model.
Interpreting the Lognormal Parameters
Because the random effect has mean zero on the log scale, setting \(\eta_{V,i}=0\) gives \(V_{\max,i}=TVV_{\max}\), and setting \(\eta_{K,i}=0\) gives \(K_{m,i}=TVK_m\). Under the stated normal assumptions, the typical value is also the median and geometric mean of the corresponding individual parameter distribution. It is not generally the arithmetic mean: the exponential transformation makes the distribution right-skewed.
If \(\eta\sim N(0,\omega^2)\) and \(\theta=TV\exp(\eta)\), then \(\log(\theta)=\log(TV)+\eta\). The normal moment-generating function gives \(E[\exp(\eta)]=\exp(\omega^2/2)\), so
The coefficient of variation (standard deviation divided by arithmetic mean) for this lognormal distribution is \(\sqrt{\exp(\omega^2)-1}\). Thus, \(\omega^2\) is a variance on the log scale, not a variance in the parameter’s original units and not a percentage variability by itself.
Worked Example: Converting Random Effects to Individual Parameters
Suppose a hypothetical population model has \(TVV_{\max}=120\) mg/h and \(TVK_m=6\) mg/L. For one hypothetical subject, let \(\eta_{V,i}=0.200\) and \(\eta_{K,i}=-0.100\). Calculate that subject’s parameter values.
For \(V_{\max}\), substitute the values into the exponential model:
For \(K_m\):
This subject has a \(V_{\max}\) above the typical value and a \(K_m\) below it. The calculation does not mean that the parameters are measured without error; it illustrates the individual parameter values implied by the specified random effects.
How the Effects Change Elimination
At concentration \(C\), the Michaelis–Menten elimination rate for subject \(i\) is
Here, \(C\) and \(K_{m,i}\) must use the same concentration units, and \(R_i(C)\) has the units of \(V_{\max,i}\), such as mg/h. At a fixed positive concentration, a higher \(V_{\max,i}\) increases the rate, while a higher \(K_{m,i}\) decreases it. The same parameter differences do not have identical consequences at every concentration.
Substituting the random-effects model into the rate equation and comparing the individual rate with the rate at the typical parameters gives
This expression shows why the net effect cannot usually be summarized by looking at only one random effect. At low concentrations, \(C\) is small relative to \(K_m\), so the rate is approximately \((V_{\max}/K_m)C\). At high concentrations, \(C\) is large relative to \(K_m\), so the rate approaches \(V_{\max}\). The following limiting argument makes the two roles explicit.
Worked Example: Comparing Elimination Rates at One Concentration
Use the hypothetical typical values \(TVV_{\max}=120\) mg/h and \(TVK_m=6\) mg/L. Compare the typical elimination rate at \(C=3\) mg/L with the rate for the subject in the previous example, whose individual parameters were \(146.6\) mg/h and \(5.429\) mg/L. Use the unrounded parameter calculations in the rate equation.
At the typical parameters:
For the individual, using \(V_{\max,i}=146.568\) mg/h and \(K_{m,i}=5.42902\) mg/L before rounding:
At this concentration, the individual’s rate is about \(52.17-40.00=12.17\) mg/h greater than the typical-parameter rate. Both random effects matter: the higher \(V_{\max}\) raises the rate, and the lower \(K_m\) also raises it. This is a comparison at the same concentration, not a claim that these subjects would necessarily have that concentration at the same time after a dose.
Worked Example: Opposing Effects Can Reduce the Rate
Consider a second hypothetical subject with the same typical values, but with \(\eta_{V,i}=-0.150\) and \(\eta_{K,i}=0.250\). First derive the individual parameters, then calculate the elimination rate at \(C=3\) mg/L.
The individual parameters are
Using the unrounded values in the rate calculation:
The rate is below the typical rate of \(40.00\) mg/h at the same concentration. Both effects act in that direction here: the lower \(V_{\max}\) reduces the rate, and the higher \(K_m\) reduces it further. If the effects had opposite signs, their contributions could partly offset; the rate equation, rather than either random effect alone, determines the net result.
Variability, Assumptions, and Checks
The lognormal specification is useful when parameters must remain positive and proportional variation is more plausible than additive variation. It makes multiplicative changes natural: an effect of \(0.200\) multiplies the typical parameter by \(\exp(0.200)\), rather than adding \(0.200\) units to a parameter. The model nevertheless makes assumptions that should be considered in context:
- The typical values are positive, and individual values are generated by the stated exponential relationships.
- The random effects are normally distributed with mean zero and the specified variances on the log scale.
- For the basic formulation here, the two random effects are independent. Independence is a modeling assumption, not a consequence of using lognormal parameters.
- The random effects represent the modeled between-subject variation; they do not replace a residual-error model for observations.
A useful scale check is to calculate the arithmetic means and coefficients of variation implied by the estimated variances. For example, if \(\omega_V^2=0.09\) and \(\omega_K^2=0.25\), with typical values of \(120\) mg/h and \(6\) mg/L, then
These arithmetic means exceed the typical values, while the medians remain \(120\) mg/h and \(6\) mg/L. The implied coefficients of variation are \(\sqrt{\exp(0.09)-1}=0.3069\), or about \(30.69\%\), for \(V_{\max}\), and \(\sqrt{\exp(0.25)-1}=0.5329\), or about \(53.29\%\), for \(K_m\).
In practice, individual parameter estimates and residual patterns can be examined for implausible values or systematic discrepancies that suggest the model does not describe the data adequately. A random-effects distribution that poorly represents the estimated subject effects is another warning sign. Interpretation also depends on the concentration range represented in the data: as the earlier material on identifiability of \(V_{\max}\) and \(K_m\) across concentration ranges emphasizes, limited information about the transition through saturation can make the two parameters difficult to distinguish. In that setting, apparently precise claims about their separate variability may not be justified.
Common Pitfalls and Limits
- Treating the effects as additive. Writing \(V_{\max,i}=TVV_{\max}+\eta_{V,i}\) would give the effect the units of \(V_{\max}\) and could produce negative values. In the lognormal model, the effect is dimensionless and multiplies the typical value through \(\exp(\eta_{V,i})\).
- Assuming a positive \(K_m\) effect raises elimination. A positive \(\eta_{K,i}\) raises \(K_{m,i}\), which lowers the rate at a fixed concentration. It is the value of \(K_m\), not the rate, that increases.
- Calling the typical value the arithmetic mean. Under a mean-zero normal random effect on the log scale, the typical value is the median. The arithmetic mean includes the factor \(\exp(\omega^2/2)\).
- Assuming parameter effects translate to the same rate difference at every concentration. The relative contribution of \(K_m\) changes with concentration; at high concentrations, the rate approaches \(V_{\max}\).
- Interpreting model assumptions as findings. Independent random effects are an assumption in this basic specification. The model alone does not establish that biological differences in \(V_{\max}\) and \(K_m\) are truly independent.
The lognormal model is one useful way to represent positive interindividual parameters, not a guarantee that the distributions or the Michaelis–Menten structure are appropriate for every data set. It also does not remove limitations from sparse or poorly informative concentration profiles. Its main benefit is a clear, interpretable mapping from log-scale subject effects to positive individual parameters and, through the rate law, to subject-specific elimination.
Check Your Understanding
Answer the questions, then compare your reasoning with the short solutions.
- If \(\eta_{V,i}=0\), what is \(V_{\max,i}\) relative to \(TVV_{\max}\)?
- With \(TVK_m=8\) mg/L and \(\eta_{K,i}=\log(1.5)\), calculate \(K_{m,i}\).
- For \(\eta_{V,i}=0.1\) and \(\eta_{K,i}=0.1\), what is the low-concentration rate ratio relative to the typical rate per unit concentration?
- What does the individual elimination rate approach as concentration becomes very large?
Solutions. (1) \(V_{\max,i}=TVV_{\max}\exp(0)=TVV_{\max}\). (2) \(K_{m,i}=8\exp(\log(1.5))=8(1.5)=12\) mg/L. (3) The ratio is \(\exp(\eta_{V,i}-\eta_{K,i})=\exp(0)=1\), because the equal effects cancel in the low-concentration ratio. (4) \(R_i(C)\) approaches \(V_{\max,i}\), because \(C/(K_{m,i}+C)\) approaches 1.